How would I find the derivative of a unit step function? I understand that the unit impulse function will be used but I'm not sure how to use it.

I am trying to find the derivative of this:

$v(t) = u(t+1) - 2u(t) + u(t-1)$

$u(t) = 0$ when $t < 0$

$u(t) = 1$ when $t > 0$

The relationship between unit step function and impulse function:

δ(n) = u(n) - u(n-1)

$ δ(t)=du(t)/dt $

  • $\begingroup$ Continuous $\displaystyle \delta(t) = \frac{d u(t)}{dt}$ $\endgroup$ – arthur Nov 1 '16 at 0:15
  • $\begingroup$ Discrete $\displaystyle \delta[n] = u[n] - u[n-1] $ $\endgroup$ – arthur Nov 1 '16 at 0:17
  • $\begingroup$ Possible duplicate of How to prove that the derivative of Heaviside's unit step function is the Dirac delta?. $\endgroup$ – user137731 Nov 1 '16 at 0:22
  • $\begingroup$ @arthur Thanks for the clarification. I've edited the post. $\endgroup$ – zdub Nov 1 '16 at 0:24
  • $\begingroup$ @zdub did you mean for one of the Heaviside functions to be $u(t\color{red}{+}1)$? Otherwise the first and third terms are the same and can be simplified to $2u(t-1)$. $\endgroup$ – user137731 Nov 1 '16 at 0:26

The derivative of unit step $u(t)$ is Dirac delta function $\delta(t)$, since an alternative definition of the unit step is using integration of $\delta(t)$ here.




  • $\begingroup$ Thanks! So the graph of the derivative would just be vertical lines at t = -1, 0 and 1? $\endgroup$ – zdub Nov 1 '16 at 0:21
  • $\begingroup$ @zdub The way that engineers draw it, there should be vertical arrows at $1$ and $0$, each of which is $2$ units tall (but at $0$ it should point in the opposite direction). $\endgroup$ – user137731 Nov 1 '16 at 0:24
  • $\begingroup$ @zdub You have two $u(t-1)$ in your question, or $2u(t-1)$. This means the derivative is a $2\delta(t-1)$ that is represented by an upward arrow with amplitude $2$ at $t=+1$. If your actual question is $u(t-1)+2u(t)+u(t+1)$, then there will be three upward delta functions in the derivative: two of them are at $t=\pm1$ with unit scale, and the third at origin with scale $2$. $\endgroup$ – msm Nov 1 '16 at 0:33

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